Sanonymous
26.02.2020 •
Chemistry
MnO2 + 4 HCI + Cl2 + MnCl2 + 2 H20
Given 145.7 grams of manganese (IV) oxide, how much hydrochloric acid (HCI) is
needed to use up the MnO2 completely?
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Ответ:
Explanation:
i did this question before once
Ответ:
244.79g of HCl
Explanation:
MnO2 + 4HCI —> Cl2 + MnCl2 + 2H20
Molar Mass of MnO2 = 54.9 + (2x16) = 54.9 + 32 = 86.9g/mol
Molar Mass of HCl = 1 + 35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 4 x 36.5 = 146g
From the equation,
86.9g of MnO2 required 146g of HCl.
Therefore, 145.7g of MnO2 will require = (145.7 x 146) / 86.9 = 244.79g of HCl
Ответ:
Magnesium has three naturally occurring isotopes (Mg-24, Mg-25, and Mg-26). Masses and natural abundances for two isotopes are listed here. Isotope Mass (amu) Abundance (%) Mg-24 23.9850 79 Mg-25 24.9858 10 Mg-26 ? ? Find the natural abundance and atomic mass of Mg-26 . Express your answer using two significant figures. thank you
its 24 and 79